Why the spatial inversion operation $P$ in two space dimensions is $(x, y)→(−x, y)$, whereas in three space dimensions it is $(x, y, z)→(−x,−y,−z)$?

The parity operation in quantum mechanics and quantum field theory is $$\hat P|\vec r\rangle=|-\vec r\rangle$$, which we can check from the Fourier transform.

Why the spatial inversion operation $$P$$ in two space dimensions is (x, y) → (−x, y), rather than (x, y) → (−x, -y), based on this definition? Can we interpret the physical action for the 3D action as a reflection about the origin, whereas in 2D is a reflection across the $$y$$-axis (why not $$x$$-axis)? Can we read the parity transformation from the tensor $$\eta^{\mu\nu}=diag(1,-1,-1)$$?

The parity transformation needs to have a determinant of $$-1$$. In all dimensions, it is safe to define it as $$(x_1, \dots, x_d) \mapsto (-x_1, x_2, \dots, x_d)$$ where only one co-ordinate is flipped. Let's call this the good parity transformation because it generalizes well to any $$d$$.

If $$d$$ is odd, $$(x_1, \dots, x_d) \mapsto (-x_1, \dots, -x_d)$$ is also valid and this seems to be what you learned first. I will call this the bad parity transformation because in even $$d$$ it would just be a rigid rotation with determinant $$1$$.

The point is that the good and bad parity transformations in odd dimension differ by a rigid rotation. I.e. something with determinant $$1$$. This also explains why there was no loss of generality in reflecting $$x_1$$ instead of $$x_2$$ for example in the good one. So you can quotient $$O(d)$$ by any of these and get the same subgroup $$SO(d)$$ as a result.

• Thanks for the answer! Why both transformations are valid in odd dimensions? Does that mean we have two ways to define $|-\vec r\rangle$?
– IGY
Oct 22, 2023 at 15:51
• There is only one transformation that should be labelled with $\left | -\vec{r} \right >$. But there are infinitely many transformations that have a right to be called parity. The only requirement is determinant -1. Oct 22, 2023 at 16:09
• Thank you! So $|-\vec r\rangle$ is the only transformation that flips the eigenvalue, right? We can call this a parity transformation if we, in 3 dimensions, define $(x,y,z)->(-x,-y,-z)$?
– IGY
Oct 22, 2023 at 16:33
• The particular parity transformation you mention is the only one that has $-1$ as its only eigenvalue. The other ones will have eigenvalues of both $1$ and $-1$ where the multiplicity of $-1$ is odd. Oct 23, 2023 at 0:39
1. Note that there are more reflections than a point reflection, e.g. mirror reflections.

2. Presumably we are looking for an element $$P\in O(n)$$ such that $$O(n)=SO(n) \sqcup P\circ SO(n)$$, i.e. $$\det(P) =-1$$; preferably $$P$$ should be an involution.

• Thanks! Are $(x,y)->(-x,y)$ and $(x,y)->(-x,-y)$ both involution?
– IGY
Oct 22, 2023 at 15:52
• Involution means its own inverse so you can check. Oct 22, 2023 at 16:11